Tony ran laps around a track to raise money for a hospital. Tony raised $15 plus $1.50 per lap that he ran. He raised a total of $255.

Let x represent the number of laps Tony ran.

What expression completes the equation to determine the total number of laps Tony ran?

How many laps did Tony run?

Answer:

[tex]\frac{b^2}{2}[/tex]

Step-by-step explanation:

We are supposed to write an algebraic expression for given word phrase.

The square of b would be 'b' raised to second power: [tex]b^2[/tex].

Half the square of b would be [tex]b^2[/tex] divide by 2: [tex]\frac{b^2}{2}[/tex].

Therefore, our required expression would be [tex]\frac{b^2}{2}[/tex].

area of a square = S^2

to find length of side take square toot of area

sqrt(59) = 7.68114

to nearest tenth = 7.7 inches

The required speed of the current is given as 9 miles per hour.

Given that,
A boat traveled 252 miles downstream and back. The trip downstream took 12 hours. The trip back took 84 hours. What is the speed of the boat in still water is to be determined.

Speed is defined as when an object is in motion, the distance covered by that object per unit of time is called speed.

Here,

let the speed of the current be y, and the speed of the boat at still water be x,
According to the question,
Downstream speed,
252 / 12 = x + y - - - -(1)
upstream speed,
252 / 84 = x - y  - - - (2)
Adding equations 1  and 2
21  + 3 = 2x
x = 24 / 2
x = 12 miles/hour
Now,
3 = 12  - y
y = 9 miles/ hour

Thus. the required speed of the current is given as 9 miles per hour.

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Final answer:

To find the speed of the boat in still water and the current, we used the relation of speed, distance, and time for downstream and upstream travel to set up equations. We solved these to determine that the boat's speed in still water is 12 miles per hour, and the current's speed is 9 miles per hour.

Explanation:

The question asks to determine the speed of the boat in still water and the speed of the current given the distances and times for traveling downstream and upstream. Using the given data, we can set up two equations based on the speed-distance-time relationship:

Downstream speed = Speed of boat in still water + Speed of current

Upstream speed = Speed of boat in still water - Speed of current

Let v represent the speed of the boat in still water, and c represent the speed of the current. We have the equations:

(v + c) = 252 miles / 12 hours = 21 miles/hour (downstream)

(v - c) = 252 miles / 84 hours = 3 miles/hour (upstream)

Adding these two equations gives us:

2v = 24 miles/hour

Dividing by 2:

v = 12 miles/hour

Substituting v in either equation to find c:

c = 21 miles/hour - v = 21 miles/hour - 12 miles/hour

c = 9 miles/hour

Therefore, the speed of the boat in still water is 12 miles/hour, and the speed of the current is 9 miles/hour.

we know that

The distance 's formula between two points is equal to

[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]

Step [tex]1[/tex]

Find the distance MN

[tex]M(-2,1)\\N(-1,4)[/tex]

Substitute in the distance's formula

[tex]d=\sqrt{(4-1)^{2}+(-1+2)^{2}}[/tex]

[tex]d=\sqrt{(3)^{2}+(1)^{2}}[/tex]

[tex]dMN=\sqrt{10}\ units[/tex]

Step [tex]2[/tex]

Find the distance NL

[tex]N(-1,4)\\L(2,4)[/tex]

Substitute in the distance's formula

[tex]d=\sqrt{(4-4)^{2}+(2+1)^{2}}[/tex]

[tex]d=\sqrt{(0)^{2}+(3)^{2}}[/tex]

[tex]dNL=3\ units[/tex]

Step [tex]3[/tex]

Find the distance LM

[tex]L(2,4)\\M(-2,1)[/tex]

Substitute in the distance's formula

[tex]d=\sqrt{(1-4)^{2}+(-2-2)^{2}}[/tex]

[tex]d=\sqrt{(-3)^{2}+(-4)^{2}}[/tex]

[tex]dLM=5\ units[/tex]

Step [tex]4[/tex]

Find the perimeter of the triangle LMN

we know that

The perimeter of a triangle is equal to the sum of the three length sides

In this problem

[tex]Perimeter=MN+NL+LM[/tex]

substitute the values in the formula

[tex]Perimeter=(\sqrt{10}+3+5)\ units[/tex]

[tex]Perimeter=(8+\sqrt{10})\ units[/tex]

therefore

the answer is the option D

the perimeter of the triangle LMN is equal to [tex](8+\sqrt{10})\ units[/tex]

Triangle DPW was translated 1 unit to the left, then it was translated 5 units down and then it was rotated by 90 degrees counterclockwise.

19.04 dollars.

The student is asking how to calculate the cost of driving 300 kilometers in a fuel-efficient car, with the cost given in dollars instead of euros. To calculate this, we need to perform several steps:

Step-by-step, this calculation is:

Therefore, it will cost approximately 19.04 dollars to drive 300 kilometers.

Answer:

[tex]8r^6s^3-5r^5s^4+r^4s^5+5r^3s^6[/tex]

Step-by-step explanation:

We want to simplify;

[tex](8r^6s^3-9r^5s^4+3r^4s^5)-(2r^4s^5-5r^3s^6-4r^5s^4)[/tex]

We expand the bracket to obtain;

[tex]8r^6s^3-9r^5s^4+3r^4s^5-2r^4s^5+5r^3s^6+4r^5s^4[/tex]

We now group the like terms to obtain;

[tex]8r^6s^3-9r^5s^4+4r^5s^4+3r^4s^5-2r^4s^5+5r^3s^6[/tex]

We now simplify to get;

[tex]8r^6s^3-5r^5s^4+r^4s^5+5r^3s^6[/tex]

The correct answer is C

Answer:

c

Step-by-step explanation:

Answer:

Option C - [tex]2\frac{1}{6}-5\frac{2}{3}=-3\frac{1}{2}[/tex]

Step-by-step explanation:

Given : Expression [tex]2\frac{1}{6}-5\frac{2}{3}[/tex]

To find : Perform the indicated operation in the expression ?

Solution :

Expression [tex]2\frac{1}{6}-5\frac{2}{3}[/tex]

Write mixed fraction into fraction,

[tex]2\frac{1}{6}-5\frac{2}{3}=\frac{13}{6}-\frac{17}{3}[/tex]

Taking Least common denominator,

[tex]2\frac{1}{6}-5\frac{2}{3}=\frac{13-34}{6}[/tex]

[tex]2\frac{1}{6}-5\frac{2}{3}=-\frac{21}{6}[/tex]

[tex]2\frac{1}{6}-5\frac{2}{3}=-\frac{7}{2}[/tex]

Write improper fraction into mixed fraction,

[tex]2\frac{1}{6}-5\frac{2}{3}=-3\frac{1}{2}[/tex]

Therefore, Option C is correct.

Proportional relationships are relationships with equal ratio. Mandy's conclusion is incorrect because the relationship is proportional, and it has a uniform ratio of 5.

To determine if the input and output are proportional, we simply divide the output (y) by the corresponding input (x).

i.e.

[tex]Ratio = \frac yx[/tex]

So, we have:

[tex]Ratio = \frac 51 = 5[/tex]

[tex]Ratio = \frac{10}{2} = 5[/tex]

[tex]Ratio = \frac{25}{5} = 5[/tex]

[tex]Ratio = \frac{50}{10} = 5[/tex]

For the four input and output data, the ratios are equal (i.e. 5).

This means that the relationship is proportional.

Hence, Mandy's conclusion is incorrect

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Mandy is correct about the proportional relationship as all the ratios of y to x in her table are equivalent to 5/1. However, there is an error with the last ratio listed as 10/50, which should be 50/10 to match the information in the table. After correcting this typo, it is clear that the relationship is proportional because all of the ratios simplify to the same value.

Mandy is indeed correct. For a set of ratios to represent a proportional relationship, they all must have the same value when simplified. From Mandy's table, the ratios formed from the inputs (x) and outputs (y) are 5/1, 10/2, 25/5, and 50/10. When simplified, these should give us a constant ratio if the relationship is proportional. Simplifying the ratios, we get:

Since all ratios simplify to the same number, 5/1, it indicates that the relationship between x and y is indeed proportional. However, the last ratio listed in the question, 10/50, is incorrect as it should be 50/10 to maintain consistency with the rest of the table. If we correct this ratio to 50/10, then all ratios confirm a proportional relationship.

To further validate this, a proportion can be created by setting two ratios equal to each other. For example, we can compare the first and the second ratio: 5/1 = 10/2. When you cross-multiply, the resulting equations, 5*2 = 10*1, are true, again confirming proportionality.

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Answer: Hence, option 'A', 'B', 'D' , 'E' are correct.

Step-by-step explanation:

There are many different types of variations  as follows:

1) Direct variations in which the relationship between two different variables is direct.

2) Inverse variations in which the relationship between two different variables is opposite.

3) Joint variations in which the relationship between one variable varies directly with the product of two or more variables.

4) Combined variations in which one variable is directly proportion to second variable but inversely proportion with third variables.

Hence, option 'A', 'B', 'D' , 'E' are correct.

The expression for the 1 less than the quotient of a number n and 6 is n/6 - 1.

It is defined as the combination of constants and variables with mathematical operators.

It is given that:

The word expression:

1 less than the quotient of a number n and 6:

Here n is the number:

A number is a mathematical entity that can be used to count, measure, or name things. For example, 1, 2, 56 etc. are the numbers.

The quotient of a number n and 6 = n/6

1 less than the quotient of a number n and 6:

n/6 - 1

If in the linear expression, one variable is present, then the expression is known as the linear expression in one variable.

Thus, the expression for the 1 less than the quotient of a number n and 6 is n/6 - 1.

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Answer: The value of x = 5

Step-by-step explanation:

Given : ∠E and ∠F are vertical angles  m∠E = 8x + 8 and m∠F = 2x + 38.

We know that the vertical angles are congruent.

It means [tex]\angle{E}\cong\angle{F}[/tex]

[tex]\Rightarrow\ m\angle{E}=\ m\angle{F}[/tex]

[tex]\Rightarrow\ 8x+8=2x+38[/tex]

Subtract 2x and 8 from both sides , we get

[tex]6x=30[/tex]

Divide both sides by 6 , we get

[tex]x=5[/tex]

Hence, the value of x = 5

Answer:

  y = (-x -6)/2

Step-by-step explanation:

You want to solve x +2y = -6 for y.

The first step here is to get the y-term by itself on one side of the equal sign. We can do that by subtracting x from both sides of the equation.

  x +2y -x = -6 -x

  2y = -6 -x

Now, we divide by the coefficient of y to get y by itself.

  2y/2 = (-6 -x)/2

  y = (-x -6)/2

We can also write this as ...

  y = -x/2 -3

Answer:

The expression represents [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] is [tex]\( y = \frac{-6 - x}{2} \)[/tex]

Explanation:

To solve the equation [tex]\(x + 2y = -6\)[/tex] for [tex]\(y\)[/tex], follow these steps:

1. Subtract [tex]\(x\)[/tex] from both sides of the equation to isolate the term containing [tex]\(y\)[/tex]:

[tex]\[2y = -6 - x\][/tex]

2. Divide both sides of the equation by [tex]\(2\)[/tex] to solve for [tex]\(y\):[/tex]

[tex]\[y = \frac{-6 - x}{2}\][/tex]

So, the solution for [tex]\(y\) is \(y = \frac{-6 - x}{2}\).[/tex]